Vinculum

He didn't capitalise the letter p in this name lol

I'm pretty sure CPhill would say "CPhill" instead of "Cphill"

Also, CPhill never puts his name at the end of his posts just the  

Edit: I double checked in Wolfram|Alpha.

It should be right.

https://www.wolframalpha.com/input?i2d=true&i=Power%5B%5Cleft%5C%2840%292y-7%5Cright%5C%2841%29%2C5%5D

Sorry Bro lemme check again

You mean 3+7+14+50? not 3+1+14+50...

-Vinculum

Edit : 3+7+14+50

its 74

Hello, 

\(x^2-y^2=51+xy\)

\(x^2-y^2-xy=51\)

\(x^2-yx-y^2-51=0\)

Solve with the quadratic formula:

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\cdot \:1\cdot \left(-y^2-51\right)}}{2\cdot \:1}\)

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{5y^2+204}}{2\cdot \:1}\).

I think you got \(y\)...

Let me know if you need help becoz this might get very messy.Good luck

f(z) = z^2 if z is real

f(z) = z + 1 if z is not real

f(i): i+1

f(1): 1^2 = 1

f(-1): -1^2 = -1

f(-i): -i+1

So we got...

(i+1)+1+(-1)+(-i+1) = 2

-Vinculum

Do you think I'm scared guest????

Think again !!!!

-Vinculum

f(x) = 2 - x if x < 1

f(x) = 2x - x^2 if x >= 1

f(-3): 2+3 = 5

f(0): 2-0 = 2

f(3): 2(3)-(3)^2 = -3

5+2-3 = 4

-Vinculum

The answer should be about: yˆ=−0.002x^2−2.051x+31.179

-Vinculum